Does there exist a $c: P(X<c)=2P(X>c)$

Question

The proportion of impurities in a certain compound is a random variable with density $f(x) = k x (1-x)$, where x belongs to $[0,2]$.

1. Calculate $k$.

2. Find the accumulated function.

3. Find $c$, a constant, such that $P(X < c)= 2 P(X>c)$.

4. Calculate the density of $Y=2/X$.

I found $k=3$ and accumulated function is $x^2-x$

But task 3 seems not to be possible for me. I've tried but can't reach any solution.

Or maybe the function I've found is wrong? Anybody knows?

Answer 1

Let $p=P(X<c)$. Then the condition: $\exists c:P(X<c) = 2P(X>c)$ is the same as saying:

$$\exists p \in [0,1]: p = 2(1-p) = 2 - 2p \implies 3p=2 \implies p=\frac{2}{3}$$

So you're looking for the 66th percentile of your variable.